My friend asked me this question as a joke but I'm genuinely curious as to what the answer is. What is $$e^{\scriptscriptstyle i^{e^{i^{.^{.^{.}}}}}}$$?
My attempt:
I first set $z$ equal to $e^{\scriptscriptstyle i^{e^{i^{.^{.^{.}}}}}}$.
$z=e^{\scriptscriptstyle i^{e^{i^{.^{.^{.}}}}}}$
$z=e^{\scriptscriptstyle i^z}$
$z=e^{iz}$
I then used the Euler's Identity $e^{i\pi}=-1$ to find that $e^i=(-1)^\frac{1}{\pi}$
$z=(-1)^{\frac{z}{\pi}}$
$z=i^{\frac{2z}{\pi}}$
I really don't know what to do at this point. Is this even solvable? Thanks in advance.
Defining $$f_1=e^i \qquad \text{and} \qquad f_n=e^{i^{f_{n-1}}}$$ I do not knwow if we could find a closed form (I cannot !) but we can compute and get the following results (from a CAS, for sure) $$\left( \begin{array}{cc} n & f_n \\ 10 & 0.8998004087100258109381994\,+\,0.5136686926540903864487995 \,i \\ 20 & 0.9391912394435662668247780\,+\,0.4748699491196762314648294 \,i \\ 30 & 0.9347741922806840376959014\,+\,0.4760068686818573570569700 \,i \\ 40 & 0.9351388632350392268316969\,+\,0.4761191810522562585293457 \,i \\ 50 & 0.9351177256059159924314937\,+\,0.4760952592226891475767457 \,i \\ 60 & 0.9351181910018841373863410\,+\,0.4760978887346746827736599 \,i \\ 70 & 0.9351182724741915050552121\,+\,0.4760976807368583340653701 \,i \\ 80 & 0.9351182575862408277286842\,+\,0.4760976920304049178539746 \,i \\ 90 & 0.9351182591415594580823332\,+\,0.4760976918736317087490291 \,i \\ 100 & 0.9351182590237006433692816\,+\,0.4760976918169832534042577 \,i \\ 110 & 0.9351182590296427005539796\,+\,0.4760976918261676061832348 \,i \\ 120 & 0.9351182590296191471242818\,+\,0.4760976918252528368153811 \,i \end{array} \right)$$
They stabilize quite fast. The problem is that the last numbers are not recognized by inverse symbolic calculators.